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## Q. 1.3

Find the gravitational force exerted by the earth on a 70-kg man whose elevation above the surface of the earth equals the radius of the earth. The mass and radius of the earth are $M_{e} = 5.9742 × 10^{24}\:kg$ and $R_{e} = 6378\:km$, respectively.

## Verified Solution

Consider a body of mass m located at the distance $2R_{e}$ from the center of the earth (of mass $M_{e}$). The law of universal gravitation, from Eq. (11.4), states that the body is attracted to the earth by the force F given by

$F=G\frac{mM_{e}}{(2R_{e})^{2}}$

where $G = 6.67 × 10^{−11}\:m^{3}/(kg·s^{2})$ is the universal gravitational constant. Substituting the values for G and the given parameters, the earth’s gravitational force acting on the 70-kg man is

$F=(6.67\times 10^{-11})\frac{(70)(5.9742\times10^{24})}{[2(6378\times 10^{3})]^{2}}=171.4\:N$